acm.jlu.edu.cn-2098-All in All

博客介绍了一种新的加密技术,需编写程序验证消息是否真的编码在最终字符串中。给定两个字符串 s 和 t,判断 s 是否为 t 的子序列。还给出了输入输出规范、示例,以及一段较易理解但速度一般的实现代码。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

All in All



You have devised a new encryption technique which encodes a message by inserting between its
characters randomly generated strings in a clever way. Because of pending patent issues we will
not discuss in detail how the strings are generated and inserted into the original message. To
validate your method, however, it is necessary to write a program that checks if the message is
really encoded in the final string.



Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can
remove characters from t such that the concatenation of the remaining characters is s.

 

Input Specification

The input contains several testcases. Each is specified by two strings s, t of alphanumeric
ASCII characters separated by whitespace. Input is terminated by EOF.

 

Output Specification

For each test case output, if s is a subsequence of t.

 

Sample Input

sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter

 

Sample Output

Yes
No
Yes
No


此代码速度一般 但较易理解

#include<iostream> #include<cstring> using namespace std; void main() { char s[100]; while(cin>>s) { int len=strlen(s); char ch; cin>>ch; int idx=0; while(ch!='/n') { if(ch==s[idx]) idx++; cin.get(ch); } if(idx==len) cout<<"Yes"<<endl; else cout<<"No"<<endl; } }

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值